Problem 18 - College Algebra

Question

Solve each rational equation for x. State all x-values that are excluded from the solution set. \(\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{(x-1)(x-2)}\)

Step-by-Step Solution

Answer

The solution is \(x = 4\) with excluded values \(x = 1\) and \(x = 2\).

Step 1: Identify the Least Common Denominator (LCD)

The given equation is \( \frac{3}{x-2} = \frac{1}{x-1} + \frac{7}{(x-1)(x-2)} \). To eliminate the fractions, first determine the least common denominator (LCD). The LCD here is \((x-1)(x-2)\).

Step 2: Eliminate the Fractions

Multiply every term in the equation by \((x-1)(x-2)\) to clear the fractions:\[ (x-1)(x-2) \cdot \frac{3}{x-2} = (x-1)(x-2) \cdot \frac{1}{x-1} + (x-1)(x-2) \cdot \frac{7}{(x-1)(x-2)} \].This simplifies to:\[ 3(x-1) = (x-2) + 7 \].

Step 3: Simplify the Equation

Expand and simplify both sides of the equation:On the left side: \( 3(x-1) = 3x - 3 \).On the right side: \((x-2) + 7 = x + 5 \).The equation becomes:\[ 3x - 3 = x + 5 \].

Step 4: Solve the Linear Equation

Rearrange terms to isolate \(x\):Subtract \(x\) from both sides:\[ 3x - x - 3 = 5 \]\[ 2x - 3 = 5 \].Add 3 to both sides:\[ 2x = 8 \].Divide both sides by 2:\[ x = 4 \].

Step 5: Determine Excluded Values

Identify the values of \(x\) that are excluded because they make the denominators zero. From the denominators \((x-1)(x-2)\), the excluded values are:- \(x = 1\)- \(x = 2\)

Step 6: Verify the Solution

The solution found is \(x = 4\). Check if \(x = 4\) is not excluded and satisfies the original equation. Substitute \(x = 4\) back into the original equation to ensure it holds true. Since \(x = 4\) does not make any denominator zero and satisfies the equation, it is a valid solution.

Key Concepts

Understanding the Least Common Denominator (LCD)

When solving rational equations, we often encounter fractions that complicate our calculations. By identifying the least common denominator (LCD), we can simplify the equation by eliminating these fractions.

The LCD is the smallest expression that both denominators can divide into without leaving a remainder. In our exercise, the equation is \( \frac{3}{x-2} = \frac{1}{x-1} + \frac{7}{(x-1)(x-2)} \).

Here, the denominators are \((x-2)\), \((x-1)\), and \((x-1)(x-2)\). The LCD becomes \((x-1)(x-2)\) because it encompasses all terms needed for each denominator. Once identified, multiplying each term by the LCD lets us eliminate the fractions, streamlining our path to the solution.

Discovering Excluded Values

Excluded values are critical in rational equations because they ensure that no invalid operations, such as division by zero, occur.

They are specific \(x\)-values that make any denominator zero, thus they must be determined and excluded from the potential solutions.

Looking at our rational equation, \((x-1)(x-2)\), we set each factor equal to zero:

• \(x-1 = 0\), gives \(x = 1\)
• \(x-2 = 0\), gives \(x = 2\)

These are the excluded values. It is essential to remember them during the solving process to avoid using a solution that might cause a mathematical error.

Linear Equation Solving Made Simple

After eliminating fractions using the LCD, rational equations often transform into linear equations that are simpler to solve. In our example, we arrive at a standard linear equation:

\[ 3x - 3 = x + 5 \].

**Steps to Solve the Linear Equation:**

• First, rearrange terms to isolate \(x\). Subtract \(x\) from both sides to get \(2x - 3 = 5\).
• Add 3 to both sides to have \(2x = 8\).
• Finally, divide each side by 2 to find \(x = 4\).

This process gives us the solution \(x = 4\), ensuring it does not match any of the previously identified excluded values, thus verifying it as a legitimate solution.